Will give brainliest if correct
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Answer:
Step-by-step explanation:
[tex]|\angle AEC|=(4x)^o \quad\implies\quad |\angle AED|=180^o-(4x)^o\\\\\\\widehat{AC}=120^o\quad\implies\quad |\angle AOC|=120^o\\\\|\angle AOC|=120^o \quad\implies\quad |\angle ADC|=\frac12\cdot120^o=60^o\\\\\\\widehat{BD}=(2x)^o\quad\implies\quad |\angle BOD|=(2x)^o\\\\|\angle BOD|=(2x)^o \quad\implies\quad |\angle ADC|=\frac12\cdot(2x)^o=x^o\\\\\\ \Delta ADE:\\\\180^o-(4x)^o+60^o+x^o=180^o\\\\-(3x)^o=-60^o\\\\x=20^o\\\\\\|\angle AED|=180^o-4\cdot20^o=100^o[/tex]